Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Software » comp.soft-sys.math.mathematica

Topic: R: R: Re: Difficult antiderivative
Replies: 2   Last Post: Dec 5, 2012 3:10 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View  
Brambilla Roberto Luigi (RSE)

Posts: 25
Registered: 2/21/12
R: R: Re: Difficult antiderivative
Posted: Dec 5, 2012 3:10 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Dear Roland,

given the integral

(1) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x] (0<r<a)

Applying your substitution x->r Cos[u] I have found (p=a/r<1)

(2) Integrate[ ArcCosh[ p Sec[u] ],u]


With your previous substitution x-> a/Cosh[u] I have found

(3) -p(NIntegrate[u Tanh[u]/Sqrt[Cosh[u]^2 -p^2],u]

And verified that (1),(2) and (3) give the same results when numerically eva
luated on a finite interval (b,r) with (b<r<a). Unfortunately none of them
seems to be solvable in the realm (rather large!) of the Mathematica8 funct
ions.
May be one has to use higher order trascendent functions? Or, wisely, stop h
ere?
Many thanks again for your attention, Rob


-----Messaggio originale-----
Da: Franzius, R. [mailto:roland.franzius@uos.de]
Inviato: marted=EC 4 dicembre 2012 11.13
A: Brambilla Roberto Luigi (RSE)
Oggetto: Re: R: Re: Difficult antiderivative

Am 30.11.2012 14:45, schrieb Brambilla Roberto Luigi (RSE):
> Dear Roland,
>
> Many thanks for your suggestion, i.e. given the indefinite integral (antid

erivative) with r<a :
>
> 1) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x]
>
> (that Mathematica8 can't solve) change the variable x->a/Cosh[u].
> Doing the substitution the integral becomes
>
> 2) Integrate[ u Tanh[u]/Sqrt[q^2 Cosh[u]^2-1],x]
>
> where q=r/a (r<a). Unfortunately also this integral is unsolvable by
> Mathematica8 (unless q=1).
>
> Alexei Boulbitch wrote (29 november)
> ....
> most of indefinite integrals have this property ("does not exist") , and o

nly smaller part of them can be expressed in terms of analytical and special
functions.
> ....
> For some integrals you can find the solution, for others you cannot, and n

o general rule exists that would help you to distinguish one group from the
other.
>
> Are (1) and (2) cases of this unhappy class?
>


There was probably a typo somewhere.

I didn't store the procedure and cam't reproduce the complete polylog
results.

The most simple algebraic form may be

replace
Sqrt(r^2-x^2) /. x->r Cos[u]
to get a Sin[u] which is cancled by the derivative

So you are left with

Integrate[ ArcCosh[ p Csc[u] ], u ]

This function is a bit strange, the Argument of ArcCosh has to be >1
demanding

0 < Cos[u] < Min[1,p]


Apply TrigToExp and with some manipulations on the Log you get a term

Integrate[Log[Cos[u]],u ]

an insolvable rest which my look like

Integrate[ Log[1 + Sqrt[ 1- q^2 Cos[u]^2] ], u ]


Regards

Roland Franzius

RSE SpA ha adottato il Modello Organizzativo ai sensi del D.Lgs.231/2001, in
forza del quale l'assunzione di obbligazioni da parte della Societ=E0 avvie
ne con firma di un procuratore, munito di idonei poteri.
RSE adopts a Compliance Programme under the Italian Law (D.Lgs.231/2001). Ac
cording to this RSE Compliance Programme, any commitment of RSE is taken by
the signature of one Representative granted by a proper Power of Attorney.
Le informazioni contenute in questo messaggio di posta elettronica sono ris
ervate e confidenziali e ne e' vietata la diffusione in qualsiasi modo o for
ma. Qualora Lei non fosse la persona destinataria del presente messaggio, La
invitiamo a non diffonderlo e ad eliminarlo, dandone gentilmente comunicazi
one al mittente. The information included in this e-mail and any attachments
are confidential and may also be privileged. If you are not the correct rec
ipient, you are kindly requested to notify the sender immediately, to cancel
it and not to disclose the contents to any other person.




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.